Problem 48 - Calculus Early Transcendentals: Pearson New International Edition

Question

Many software packages have programs for calculating limits, although you should be warned that they are not infallible. To develop confidence in your program, use it to recalculate some of the limits in Problems 1-28. Then for each of the following, find the limit or state that it does not exist. $$ \lim _{x \rightarrow 0^{+}} x^{x} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Rewriting the Expression

The original limit to evaluate is \( \lim _{x \rightarrow 0^{+}} x^{x} \). Rewriting \( x^{x} \) in an exponential form using natural logarithms gives us \( x^{x} = e^{x \ln x} \).

Step 2: Simplifying the Exponent

Now, focus on the exponent \( x \ln x \). We need to find \( \lim_{x \rightarrow 0^{+}} x \ln x \). As \( x \to 0^{+} \), \( \ln x \) tends to \(-\infty\). Therefore, investigate the product \( x \ln x \).

Step 3: Changing to an Indeterminate Form

Rewrite \( x \ln x \) as \( \frac{\ln x}{1/x} \). As \( x \to 0^{+} \), this is an indeterminate form \( \frac{-\infty}{\infty} \). This allows us to use L'Hopital's rule to find the limit.

Step 4: Apply L'Hopital's Rule

Differentiate the numerator and the denominator separately. The derivative of \( \ln x \) is \( \frac{1}{x} \), and the derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \). Applying L'Hopital's Rule gives us: \( \lim_{x \rightarrow 0^{+}} \frac{\ln x}{1/x} = \lim_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \rightarrow 0^{+}} -x = 0 \).

Step 5: Final Evaluation of the Original Limit

With the exponent \( \lim_{x \rightarrow 0^{+}} x \ln x = 0 \), we have \( \lim_{x \rightarrow 0^{+}} x^{x} = e^{0} = 1 \). Thus, the limit exists and it equals 1.

Key Concepts

L'Hopital's Rule

When dealing with indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hopital's Rule is an invaluable tool. This rule provides a way to evaluate limits more straightforwardly by differentiating the numerator and denominator separately. Once differentiated, you can reevaluate the limit.
For instance, in the original problem, the expression \( \frac{\ln x}{1/x} \) forms an indeterminate \( \frac{-\infty}{\infty} \) as \( x \to 0^+ \). Applying L'Hopital's Rule by differentiating gives us \( \frac{\frac{1}{x}}{-\frac{1}{x^2}} \).

• This simplifies to \( -x \), which, upon taking the limit as \( x \to 0^+ \), results in 0.
• Always remember, L'Hopital's Rule requires that the initial expression must be an indeterminate form before applying the rule.
• Ensure the functions involved are differentiable in the neighborhood of the point under evaluation.

Natural Logarithms

Natural logarithms, denoted as \( \ln(x) \), are logarithms to the base \( e \), where \( e \approx 2.71828 \). They play a critical role in calculus due to their properties and ease of differentiability.
In the context of the original problem, expressing \( x^x \) as \( e^{x \ln x} \) is key. This transformation utilizes the fact that \( a^b = e^{b \ln a} \), leveraging the natural logarithm to simplify exponents.

• Natural logarithms help convert complex powers and roots into simpler expressions.
• The derivative of \( \ln(x) \) is \( \frac{1}{x} \), a simple yet powerful result used for limit calculations and applications of L'Hopital's Rule.
• When approaching limits, it's important to recognize the behavior of \( \ln(x) \) as \( x \to 0^+ \): it goes to \(-\infty\).

Indeterminate Forms

Understanding indeterminate forms is crucial for solving limits where direct substitution does not work. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), 0\( \cdot \infty \), \( \infty - \infty \), and several others.
In this exercise, \( x \ln x \) becomes \( \frac{-\infty}{\infty} \) as \( x \to 0^+ \). This is a classic case for using L'Hopital's Rule to simplify and find the limit.

• Indeterminate forms indicate more work is needed beyond simple substitution to determine the limit.
• L'Hopital's Rule, algebraic manipulation, or series expansion might be necessary to resolve the limit.
• Identifying whether a limit is in an indeterminate form helps in choosing the proper technique to evaluate it.

Handling indeterminate forms often requires creativity and a solid understanding of calculus principles, as seen in steps like rewriting \( x \ln x \) into a simpler ratio.